Find the locus of the middle points of chords of the circle $x^{2} + y^{2} = a^{2}$ which subtend a right angle at the point $(c, 0)$ .

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Solution

Let mid point be $(h, k)$ .

Let the on end of the chord be $(a cos θ, a sin θ)$

Other point must be $(2 h - a cos θ, 2 k - a sin θ)$ .

$(∵ M i d p o i n t x = \frac{x_{1} + x_{2}}{2}, M i d y = \frac{y_{1} + y_{2}}{2})$ .

$\Rightarrow (2 h - a cos θ)^{2} + (2 k - a sin θ)^{2} = a^{2}$

$\Rightarrow 4 h^{2} - 4 a h c o s θ + a^{2} {cos}^{2} θ + 4 k^{2} - 4 a k sin θ + a^{2} {sin}^{2} θ = a^{2}$

$\Rightarrow 4 h^{2} - 4 a h cos θ + a^{2} + 4 k^{2} - 4 a k sin θ = a^{2}$

$(∵ {sin}^{2} θ + {cos}^{2} θ = 1)$

$\Rightarrow a h cos θ + a k sin θ = h^{2} + k^{2} - (1)$

$h cos θ + k sin θ = \frac{h^{2} + k^{2}}{a}$

chord subtends right angleat $(c, 0)$

$∴$ Line joining $(c, 0) & (2 h - a cos θ, a sin θ)$ and $(c_{1} o) & (2 h - a cos θ, 2 h - a sin θ)$ are perpendicular.

So, product of slopes $= - 1$

$\frac{(0 - 2 k + a sin θ)}{(c - 2 h + a cos θ)} \times \frac{(0 - a sin θ)}{(c - a cos θ)} = - 1.$

$\Rightarrow \frac{- 2 k + a sin θ}{c - 2 h + a cos θ} = \frac{a cos θ - c}{- a sin θ}$

$\Rightarrow 2 a k sin θ - a^{2} {sin}^{2} θ = a c cos θ - c^{2}$ $- 2 a h cos θ + 2 c h$ $+ a^{2} {c o s}^{2} θ - a c cos θ$

$2 a k sin θ - a^{2} ({sin}^{2} θ + {cos}^{2} θ) = a c cos θ - c^{2} + 2 c h$ $- 2 a h cos θ - a c cos θ$ .

$\Rightarrow 2 a k sin θ - a^{2} = 2 c h - c^{2} - 2 a h cos θ$

$\Rightarrow 2 a (k sin θ + h cos θ) = 2 c h + a^{2} - c^{2}$

$F r o m (1) :$

$\Rightarrow 2 a \frac{(h^{2} + k^{2})}{a} = 2 c h + a^{2} - c^{2} .$

$\Rightarrow 2 h^{2} + 2 k^{2} = 2 c h + a^{2} - c^{2} .$

$\Rightarrow 2 h^{2} + 2 k^{2} - 2 c h + (c^{2} - a^{2}) = 0$ Ans.

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